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Thread: Limit

  1. April 17th 2009, 10:05 AM #1
    Apprentice123
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    Limit

    Consider a function f that |f(x) - Ax| \leq e^{-x}, for any number A \in R. Calculate:

    lim_x{ \to +\infty} \frac{f(x) - f(1)}{x-1}
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  2. April 17th 2009, 11:41 AM #2
    running-gag
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    Hi

    \frac{|f(x) - f(1)|}{|x-1|} = \frac{|(f(x) - Ax) + (Ax - A) + (A - f(1))|}{|x-1|}

    \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax| + |Ax - A| + |A - f(1)|}{|x-1|}

    Spoiler:

    \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax|}{|x-1|} + \frac{|Ax - A|}{|x-1|} + \frac{|A - f(1)|}{|x-1|}

    \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}

    And

    \lim_{x \rightarrow +\infty} \left(\frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}\right) = |A|

    \forall A \in \mathbb{R} ~ \lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} \leq |A|

    Therefore \lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} = 0


    It would be better if someone could confirm !

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  3. April 17th 2009, 02:05 PM #3
    Apprentice123
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    Thank you men
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